Base | Representation |
---|---|
bin | 1100000111100000111010… |
… | …0101101101110010111011 |
3 | 1202011200121001000111122111 |
4 | 3001320032211231302323 |
5 | 3221241440212220443 |
6 | 44200335304233151 |
7 | 2543400066445156 |
oct | 301701645556273 |
9 | 52150531014574 |
10 | 13323233320123 |
11 | 4277395967301 |
12 | 15b216ba6b7b7 |
13 | 7584b5cc870c |
14 | 340bc2a2cc9d |
15 | 18187b89269d |
hex | c1e0e96dcbb |
13323233320123 has 2 divisors, whose sum is σ = 13323233320124. Its totient is φ = 13323233320122.
The previous prime is 13323233320097. The next prime is 13323233320141. The reversal of 13323233320123 is 32102333232331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13323233320123 - 217 = 13323233189051 is a prime.
It is a super-3 number, since 3×133232333201233 (a number of 40 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is not a weakly prime, because it can be changed into another prime (13323233330123) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6661616660061 + 6661616660062.
It is an arithmetic number, because the mean of its divisors is an integer number (6661616660062).
Almost surely, 213323233320123 is an apocalyptic number.
13323233320123 is a deficient number, since it is larger than the sum of its proper divisors (1).
13323233320123 is an equidigital number, since it uses as much as digits as its factorization.
13323233320123 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 34992, while the sum is 31.
Adding to 13323233320123 its reverse (32102333232331), we get a palindrome (45425566552454).
The spelling of 13323233320123 in words is "thirteen trillion, three hundred twenty-three billion, two hundred thirty-three million, three hundred twenty thousand, one hundred twenty-three".
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